JEE Main 2019 · Mathematics · Binomial Theorem · medium

In the expansion of (1+x)^n, the coefficients of the 5th, 6th and 7th terms are in arithmetic progression. The sum of all possible values of n is

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In the expansion of

(1+x)^n

, the coefficients of the 5th, 6th and 7th terms are in arithmetic progression. The sum of all possible values of

n

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Answer: (A) $21$

The coefficients of the 5th, 6th and 7th terms are

\binom n4,\binom n5,\binom n6

. They are in AP when

2\binom n5=\binom n4+\binom n6

. Divide by

\binom n5

\dfrac{\binom n4}{\binom n5}=\dfrac{5}{n-4}

and

\dfrac{\binom n6}{\binom n5}=\dfrac{n-5}{6}

. So

2=\dfrac5{n-4}+\dfrac{n-5}6

, which gives

12(n-4)=30+(n-4)(n-5)

, i.e.

n^2-21n+98=0

. Hence

n=7

n=14

(both satisfy

n\ge6

). The sum of the possible values is

7+14=21

, which equals the sum of the roots of the quadratic.

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